Question

А і
A car is moving at 72 km/her after covering a
distance of 100m his nelocity is 54 km/hr calculate
the acceleration of the car ?​

Answer 1

Answer:

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Answer 2

$\huge {\bold {\underline {\underline{Given : - }}}}$

• Initial velocity of car = 72km/hr
• Distance covered = 100m
• Final velocity = 54 km/hr

$\huge {\bold {\underline {\underline{To \: find : - }}}}$

• Acceleration of car

$\huge {\bold {\underline {\underline{Solution : - }}}}$

The relation between km/hr and m/s is given by ,

$\large {\underline {\bold {\boxed{ \bigstar{ | 1km.hr {}^{ - 1} = \dfrac{5}{18} \: ms {}^{ - 1} }}}}}$

We are given that ,

• $\small\rm{v = 54km.hr {}^{ - 1} = \frac{54 \times 5}{18} = 15 \: m. {s}^{ - 1}}$

• $\small\rm{ u = 72km. {hr}^{ - 1} = \dfrac{72 \times 5}{18} = 20 \: m. {s}^{ - 1} }$

• S = 100m

From Third equation of motion ,

$\large \underline{\bold {\boxed {\bigstar{ | {v}^{2} - {u}^{2} = 2as}}}}$

where ,

• v is final velocity
• u is initial velocity
• a is acceleration
• s is distance

$: \implies \: (15) {}^{2} - (20) {}^{2} = 2(a)(100) \\ \\ : \implies \: 225- 400 = 200a \\ \\ : \implies \: - 175 = 200a \\ \\ : \implies \: a = - \dfrac{ 175}{200} \\ \\ : \implies \: a = - 0.875 \: m. {s}^{ - 2}$

∴ The Acceleration of the car is -0.875 m/s²

$\huge {\bold {\underline {\underline{Additional \: info: - }}}}$

★ The first equation of motion is given by ,

$\large \underline {\bold {\boxed {\bigstar{ \: | v = u + at}}}}$

Where ,

• v is final velocity
• u is initial velocity
• a is acceleration
• t is time

★ Second equation of motion is given by ,

$\large { \underline{\bold {\boxed{ \bigstar{ \: | s = ut + \dfrac{1}{2}at {}^{2} }}}}}$

Where ,

• S is distance
• u is initial velocity
• a is acceleration
• t is time

$\huge {\bold {\underline{\underline {Note : - }}}}$

Equations of Motion are valid only in case of constant acceleration.If acceleration has negative sign it indicates the decrease in acceleration (or) deceleration.